A particle takes `n` seconds less and acquires a velocity u m/sec. higher at one place than at another place in falling through the same distance. Calculate the product of the acceleration due to gravity at these two places.

Answer

`((u)/(n))^(2)`

Explanation

`s=(1)/(2)g_(1)t_(0)^(2)` ..(i)
`s=(1)/(2)g_(2)(t_(0)-n)^(2)` …(ii)
`u_(0)=g_(1)t_(0)` ..(iii)
`u_(0)+u=g_(2)(t_(0)-n)` ..(iv)
After solving we get `g_(1)g_(2)=((u)/(n))^(2)`